= Solution
Use the usual <cardinal arithmetic> with the <axiom of choice>. The <König theorem for cardinal numbers> states that for any index set $I$ and <cardinal numbers> $\kappa_i<\lambda_i$,
$$
\boxed{\sum_{i\in I}\kappa_i<\prod_{i\in I}\lambda_i.}
$$
Represent the <cardinal numbers> as <initial ordinals>. Let $A=\{(i,\alpha):i\in I,\alpha<\kappa_i\}$ and $B=\prod_{i\in I}\lambda_i$. There is an <injection> $A\to B$: send $(i,\alpha)$ to the function whose $i$th coordinate is $\alpha+1$ and whose other coordinates are zero. The unique nonzero coordinate identifies $i$, and its value identifies $\alpha$. All values are allowed because $\alpha+1\leq\kappa_i<\lambda_i$.
For any proposed <surjection> $F:A\to B$, its $i$th row has at most $\kappa_i$ elements, so
$$
\{F(i,\alpha)(i):\alpha<\kappa_i\}\ne\lambda_i.
$$
Choose $b_i$ to be the least omitted <ordinal>. The function $b=(b_i)_{i\in I}$ belongs to $B$ and differs from $F(i,\alpha)$ at coordinate $i$, for every $(i,\alpha)\in A$. Thus $F$ is not surjective. Cardinal comparison now gives the strict inequality. If $I=\varnothing$, the conclusion is $0<1$, also valid.
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